Question

If P,Q are two points on the line $3x+4y+15=0$ such that $OP=OQ=9$, then the area of $△OPQ$ is ?

• A. $6\sqrt{2}$
• B. $9\sqrt{2}$
• C. $12\sqrt{2}$
• D. $18\sqrt{2}$

Answer 1

The correct option is D $18\sqrt{2}$

Given in $△OPQ$, $O$ is the origin and $OP=OQ=9$

Therefore,$△OPQ$ is isosceles

Let, $C$ is the midpoint of $PQ$, $OC$ is the perpendiular to the line $PQ$.

$OC=$ Length of the perpendicular from $(0,0)$ to the line $3x+4y+15=0$

$\Rightarrow OC=\frac{|0+0+15|}{\sqrt{3^2+4^2}}=\frac{15}{\sqrt{25}}=\frac{15}{5}=3$

In right angled triangle $OCP$,

${OP}^2={OC}^2+{PC}^2\Rightarrow 9^2=3^2+{PC}^2\Rightarrow {PC}^2=81-9\Rightarrow {PC}^2=72\Rightarrow PC=\sqrt{72}=6\sqrt{2}$

Since, $C$ is the midpoint of $PQ$

$\therefore PQ=2\times 6\sqrt{2}=12\sqrt{2}$

$\therefore$ Area of $△OPQ=\frac{1}{2}\times OC\times PQ=\frac{1}{2}\times 3\times 12\sqrt{2}=18\sqrt{2}squareunits.$